Solve the equation

$$\sqrt{x+2}=x.$$

The usual route is to get rid of the root by squaring both sides:

$$\begin{aligned} \sqrt{x+2}=x\quad&\Longrightarrow\quad x+2=x^2\\ &\Longrightarrow\quad (x-2)(x+1)=0\\ &\Longrightarrow\quad x=2\ \text{or}\ x=-1. \end{aligned}$$

The calculations are all correct. The answer is not. With $x=2$: $\sqrt{4}=2$, and the equation holds. With $x=-1$: $\sqrt{1}=1$, which is not $-1$. The original equation has one solution, and the method found two. Where did the intruder come from?

One equation per line

Someone solving an equation is not calculating with numbers: they are writing a sequence of equations, each obtained from the previous one. What links one line to the next is not an equals sign, it is a statement about the sentences: “if this one is true, that one is too”. Logic has two signs for this.

  • $P\Rightarrow Q$, “$P$ implies $Q$”: whenever $P$ is true, $Q$ is true as well. But $Q$ can be true without $P$ being true.
  • $P\Leftrightarrow Q$, “$P$ is equivalent to $Q$”: both are true for exactly the same values. The arrow works both ways.

When solving an equation, the difference is decisive. If every step is a $\Leftrightarrow$, the last line has exactly the same solutions as the first. If some step is only $\Rightarrow$, the last line may have extra solutions. If some step is only $\Leftarrow$, it may have fewer.

Steps that work both ways

Some steps are always reversible:

Step Why you can go back
add the same number to both sides just subtract it again
multiply both sides by a number other than zero just divide again
replace an expression by an identical one the identity works both ways
apply a one-to-one function, such as $x\mapsto x^3$ or $x\mapsto e^x$ each result comes from a single value

Others are not:

Step What can happen
square both sides gain solutions: from $a=b$ follows $a^2=b^2$, but $a^2=b^2$ also holds if $a=-b$
multiply by an expression that can be zero gain solutions: the zeros of the expression
divide by an expression that can be zero lose solutions: the zeros of the expression

In the equation at the start, squaring was a $\Rightarrow$ in disguise. The equation $x+2=x^2$ is the question “is $\sqrt{x+2}$ equal to $x$ or to $-x$?”, and $-1$ answers the second half: $\sqrt{1}=1=-(-1)$. The intruder has a name: an extraneous root. The only defence is to go back to the original equation and check.

Why books write ⇔

Chapter 2 solved $x+\frac x2=5$ like this:

$$x+\frac x2=5\iff\frac{3x}2=5\iff 3x=10\iff x=\frac{10}3.$$

The double arrows are not decoration: they say that no step multiplied by zero or squared. When the reader sees $\iff$, checking is optional; when the reader sees $\Rightarrow$, checking is mandatory. It is a notation that does the job of a warning.

In school practice, the arrows tend to disappear, and the lines are simply stacked. The student sees no difference between a safe step and a dangerous one, and learns to check the answer “because the teacher says so”, without knowing that checking is exactly what turns a chain of $\Rightarrow$ into a correct answer.

Exercise: find the extraneous root

Solve $\sqrt{2x+3}=x$. Squaring ($\Rightarrow$): $2x+3=x^2$, $x^2-2x-3=0$, $(x-3)(x+1)=0$, $x=3$ or $x=-1$. Checking: with $x=3$, $\sqrt9=3$, it works. With $x=-1$, $\sqrt1=1\neq-1$, it is extraneous. The only solution is $x=3$.

Exercise: where did the zero go?

A student solved $x^3=4x$ by dividing by $x$: $x^2=4$, $x=\pm2$. One solution was missing. The correct step is to factorise, which is a $\Leftrightarrow$: $x^3-4x=0\iff x(x^2-4)=0\iff x(x-2)(x+2)=0\iff x\in\{-2,0,2\}$. Dividing by $x$ assumed, without saying so, that $x\neq0$, and lost exactly the case $x=0$.

Chapter index · ← Equivalence: fractions, clocks and Gauss’s ≡ · The chain of equalities: how an identity is proved →